Number of Ways to Wear Different Hats to Each Other

// Accepted solution for LeetCode #1434: Number of Ways to Wear Different Hats to Each Other
class Solution {
    public int numberWays(List<List<Integer>> hats) {
        int n = hats.size();
        int m = 0;
        for (var h : hats) {
            for (int v : h) {
                m = Math.max(m, v);
            }
        }
        List<Integer>[] g = new List[m + 1];
        Arrays.setAll(g, k -> new ArrayList<>());
        for (int i = 0; i < n; ++i) {
            for (int v : hats.get(i)) {
                g[v].add(i);
            }
        }
        final int mod = (int) 1e9 + 7;
        int[][] f = new int[m + 1][1 << n];
        f[0][0] = 1;
        for (int i = 1; i <= m; ++i) {
            for (int j = 0; j < 1 << n; ++j) {
                f[i][j] = f[i - 1][j];
                for (int k : g[i]) {
                    if ((j >> k & 1) == 1) {
                        f[i][j] = (f[i][j] + f[i - 1][j ^ (1 << k)]) % mod;
                    }
                }
            }
        }
        return f[m][(1 << n) - 1];
    }
}

// Accepted solution for LeetCode #1434: Number of Ways to Wear Different Hats to Each Other
func numberWays(hats [][]int) int {
	n := len(hats)
	m := 0
	for _, h := range hats {
		m = max(m, slices.Max(h))
	}
	g := make([][]int, m+1)
	for i, h := range hats {
		for _, v := range h {
			g[v] = append(g[v], i)
		}
	}
	const mod = 1e9 + 7
	f := make([][]int, m+1)
	for i := range f {
		f[i] = make([]int, 1<<n)
	}
	f[0][0] = 1
	for i := 1; i <= m; i++ {
		for j := 0; j < 1<<n; j++ {
			f[i][j] = f[i-1][j]
			for _, k := range g[i] {
				if j>>k&1 == 1 {
					f[i][j] = (f[i][j] + f[i-1][j^(1<<k)]) % mod
				}
			}
		}
	}
	return f[m][(1<<n)-1]
}

# Accepted solution for LeetCode #1434: Number of Ways to Wear Different Hats to Each Other
class Solution:
    def numberWays(self, hats: List[List[int]]) -> int:
        g = defaultdict(list)
        for i, h in enumerate(hats):
            for v in h:
                g[v].append(i)
        mod = 10**9 + 7
        n = len(hats)
        m = max(max(h) for h in hats)
        f = [[0] * (1 << n) for _ in range(m + 1)]
        f[0][0] = 1
        for i in range(1, m + 1):
            for j in range(1 << n):
                f[i][j] = f[i - 1][j]
                for k in g[i]:
                    if j >> k & 1:
                        f[i][j] = (f[i][j] + f[i - 1][j ^ (1 << k)]) % mod
        return f[m][-1]

// Accepted solution for LeetCode #1434: Number of Ways to Wear Different Hats to Each Other
/**
 * [1434] Number of Ways to Wear Different Hats to Each Other
 *
 * There are n people and 40 types of hats labeled from 1 to 40.
 * Given a 2D integer array hats, where hats[i] is a list of all hats preferred by the i^th person.
 * Return the number of ways that the n people wear different hats to each other.
 * Since the answer may be too large, return it modulo 10^9 + 7.
 *  
 * Example 1:
 *
 * Input: hats = [[3,4],[4,5],[5]]
 * Output: 1
 * Explanation: There is only one way to choose hats given the conditions.
 * First person choose hat 3, Second person choose hat 4 and last one hat 5.
 *
 * Example 2:
 *
 * Input: hats = [[3,5,1],[3,5]]
 * Output: 4
 * Explanation: There are 4 ways to choose hats:
 * (3,5), (5,3), (1,3) and (1,5)
 *
 * Example 3:
 *
 * Input: hats = [[1,2,3,4],[1,2,3,4],[1,2,3,4],[1,2,3,4]]
 * Output: 24
 * Explanation: Each person can choose hats labeled from 1 to 4.
 * Number of Permutations of (1,2,3,4) = 24.
 *
 *  
 * Constraints:
 *
 * 	n == hats.length
 * 	1 <= n <= 10
 * 	1 <= hats[i].length <= 40
 * 	1 <= hats[i][j] <= 40
 * 	hats[i] contains a list of unique integers.
 *
 */
pub struct Solution {}

// problem: https://leetcode.com/problems/number-of-ways-to-wear-different-hats-to-each-other/
// discuss: https://leetcode.com/problems/number-of-ways-to-wear-different-hats-to-each-other/discuss/?currentPage=1&orderBy=most_votes&query=

// submission codes start here

impl Solution {
    // Credit: https://leetcode.com/problems/number-of-ways-to-wear-different-hats-to-each-other/solutions/3112072/just-a-runnable-solution/
    pub fn number_ways(hats: Vec<Vec<i32>>) -> i32 {
        let mut dp = vec![vec![-1; 1024]; 41];
        let n = hats.len();
        let mut h2p = vec![vec![]; 41];

        for (i, item) in hats.iter().enumerate().take(n) {
            for &hat in item {
                h2p[hat as usize].push(i);
            }
        }

        Self::dfs_helper(&hats, &h2p, (1 << n) - 1, 1, 0, &mut dp)
    }

    fn dfs_helper(
        hats: &Vec<Vec<i32>>,
        h2p: &Vec<Vec<usize>>,
        all_mask: i32,
        hat: usize,
        assigned_people: i32,
        dp: &mut Vec<Vec<i32>>,
    ) -> i32 {
        if assigned_people == all_mask {
            return 1;
        }

        if hat > 40 {
            return 0;
        }

        if dp[hat][assigned_people as usize] != -1 {
            return dp[hat][assigned_people as usize];
        }

        let mut answer = Self::dfs_helper(hats, h2p, all_mask, hat + 1, assigned_people, dp);

        for &p in &h2p[hat] {
            if ((assigned_people >> p) & 1) == 1 {
                continue;
            }
            answer +=
                Self::dfs_helper(hats, h2p, all_mask, hat + 1, assigned_people | (1 << p), dp);
            answer %= 1_000_000_007;
        }

        dp[hat][assigned_people as usize] = answer;

        answer
    }
}

// submission codes end

#[cfg(test)]
mod tests {
    use super::*;

    #[test]
    fn test_1434_example_1() {
        let hats = vec![vec![3, 4], vec![4, 5], vec![5]];

        let result = 1;

        assert_eq!(Solution::number_ways(hats), result);
    }

    #[test]
    fn test_1434_example_2() {
        let hats = vec![vec![3, 5, 1], vec![3, 5]];

        let result = 4;

        assert_eq!(Solution::number_ways(hats), result);
    }

    #[test]
    fn test_1434_example_3() {
        let hats = vec![
            vec![1, 2, 3, 4],
            vec![1, 2, 3, 4],
            vec![1, 2, 3, 4],
            vec![1, 2, 3, 4],
        ];

        let result = 24;

        assert_eq!(Solution::number_ways(hats), result);
    }
}

// Accepted solution for LeetCode #1434: Number of Ways to Wear Different Hats to Each Other
function numberWays(hats: number[][]): number {
    const n = hats.length;
    const m = Math.max(...hats.flat());
    const g: number[][] = Array.from({ length: m + 1 }, () => []);
    for (let i = 0; i < n; ++i) {
        for (const v of hats[i]) {
            g[v].push(i);
        }
    }
    const f: number[][] = Array.from({ length: m + 1 }, () =>
        Array.from({ length: 1 << n }, () => 0),
    );
    f[0][0] = 1;
    const mod = 1e9 + 7;
    for (let i = 1; i <= m; ++i) {
        for (let j = 0; j < 1 << n; ++j) {
            f[i][j] = f[i - 1][j];
            for (const k of g[i]) {
                if (((j >> k) & 1) === 1) {
                    f[i][j] = (f[i][j] + f[i - 1][j ^ (1 << k)]) % mod;
                }
            }
        }
    }
    return f[m][(1 << n) - 1];
}